Question

A skier with a mass of \(63 \mathrm{kg}\) starts from rest and skis down an icy (frictionless) slope that has a length of \(50 \mathrm{m}\) at an angle of \(32^{\circ}\) with respect to the horizontal. At the bottom of the slope, the path levels out and becomes horizontal, the snow becomes less icy, and the skier begins to slow down, coming to rest in a distance of \(140 \mathrm{m}\) along the horizontal path. (a) What is the speed of the skier at the bottom of the slope? (b) What is the coefficient of kinetic friction between the skier and the horizontal surface?

Short answer

Calculating the GPE gives us: \(GPE = (63 \, \mathrm{kg}) (9.81 \, \mathrm{m/s^2}) (26.60 \, \mathrm{m}) \approx 16461.26 \, \mathrm{J}\) (joules). #tag_title# Step 2: Determine Kinetic Energy at Bottom of Slope #tag_content# Since there is no friction on the slope, we can apply the conservation of mechanical energy and assume that the GPE at the top of the slope will be completely converted into kinetic energy (KE) at the bottom. Thus, KE at the bottom of the slope will be equal to the GPE at the top: \(KE = 16461.26 \, \mathrm{J}\). #tag_title# Step 3: Determine Speed at Bottom of Slope #tag_content# We can now calculate the skier's speed at the bottom of the slope using the kinetic energy formula: \(KE = \frac{1}{2}mv^2\), where \(m\) is the mass of the skier (63 kg) and \(v\) is the speed. Solving for \(v\), we get: \(v = \sqrt{\frac{2 \cdot KE}{m}} = \sqrt{\frac{2 \cdot 16461.26 \, \mathrm{J}}{63 \, \mathrm{kg}}} \approx 22.97 \, \mathrm{m/s}\) The skier's speed at the bottom of the slope is approximately \(22.97 \, \mathrm{m/s}\). #tag_title# Step 4: Determine Deceleration on Horizontal Surface #tag_content# To find the deceleration on the horizontal surface, we can use the formula: \(v^2 = u^2 + 2as\), where \(u\) is the initial speed (\(22.97 \, \mathrm{m/s}\)), \(v\) is the final speed (0 m/s, since the skier comes to rest), \(a\) is the deceleration, and \(s\) is the distance traveled on the horizontal surface (\(96 \, \mathrm{m}\)). Solving for \(a\), we get: \(a = \frac{v^2 - u^2}{2s} = \frac{0^2 - (22.97 \, \mathrm{m/s})^2}{2(96 \, \mathrm{m})} \approx -2.74 \, \mathrm{m/s^2}\) The deceleration on the horizontal surface is approximately -2.74 m/s² (negative sign indicates the direction opposite to the motion). #tag_title# Step 5: Determine Coefficient of Kinetic Friction #tag_content# We can now use Newton's second law of motion to find the friction force acting on the skier: \(F_f = ma\), where \(F_f\) is the friction force, \(m\) is the skier's mass (63 kg), and \(a\) is the deceleration (-2.74 m/s²). The friction force is: \(F_f = (63 \, \mathrm{kg})(-2.74 \, \mathrm{m/s^2}) = -172.62 \, \mathrm{N}\) (negative sign indicates the direction opposite to the motion). Next, we use the formula for kinetic friction: \(F_f = \mu F_n\), where \(F_n\) is the normal force (equivalent to the weight of the skier in this case) and \(\mu\) is the coefficient of kinetic friction. We have: \(-172.62 \, \mathrm{N} = \mu (mg)\) Solving for \(\mu\), we get: \(\mu = \frac{-172.62 \, \mathrm{N}}{(63 \, \mathrm{kg})(9.81 \, \mathrm{m/s^2})} \approx 0.28\) The coefficient of kinetic friction between the skier and the horizontal surface is approximately 0.28.

Step-by-step solution

Determine Gravitational Potential Energy at Top of Slope

To do this, we need to determine the vertical height (\(h\)) of the slope. From the problem statement, we have the slope angle (\(32^{\circ}\)) and slope length (\(50 \, \mathrm{m}\)). We can use the sine function to find the height: \(h = L \sin{\theta} = 50 \sin{32^{\circ}} \approx 26.60 \, \mathrm{m}\). Next, we calculate the gravitational potential energy (\(GPE\)) at the top of the slope: \(GPE = mgh\), where \(m\) is the mass of the skier (\(63 \, \mathrm{kg}\)), \(g\) is the gravitational acceleration (\(9.81 \, \mathrm{m/s^2}\)), and \(h\) is the vertical height (\(26.60 \, \mathrm{m}\)).